Help to prove $\frac{1}{4a^2-8ac+13c^2}+\frac{1}{4b^2-8bc+13c^2}\ge \frac{2}{(a+b+c)^2}$

Question

If $a\ge b>c\ge 0,$ then prove $$\frac{1}{4a^2-8ac+13c^2}+\frac{1}{4b^2-8bc+13c^2}\ge \frac{2}{(a+b+c)^2}.$$ When does equality hold ? I tried to use Cauchy-Schwarz $$(a+b+c)^2\geq 2(a^2+b^2)+13c^2-4c(a+b)$$ $$\iff 6c(a+b)+ 2ab\ge a^2+b^2+12c^2,$$which is reverse when $c=0.$

I was stuck to com up with an idea using the given condition.

Hope to see some usful helps. Thank you.

Answer 1

Proof. We'll rewrite the original inequality as $$\frac{1}{4x^2+z^2}+\frac{1}{4y^2+z^2}\ge\frac{2}{(x+y+z)^2},$$where $x=a-c\ge 0; y=b-c\ge0; z=3c\ge 0\implies x+y+z=a+b+c.$

Notice that $4x^2+z^2=(2x+z)^2-4xz\le (2x+z)^2; 4y^2+z^2=(2y+z)^2-4yz\le (2y+z)^2.$

Thus, it's enough to prove that$$\frac{1}{(2x+z)^2}+\frac{1}{(2y+z)^2}\ge \frac{2}{(x+y+z)^2},$$which is easy by Cauchy-Schwarz$$LHS\ge \frac{1}{2}\left(\frac{1}{2x+z}+\frac{1}{2y+z}\right)^2\ge \frac{8}{2(x+y+z)^2}=RHS.$$ The proof is done. Equality holds at $c=0; a=b>0.$

Answer 2

Let $b=c+u$ and $a=c+u+v$.

Thus, we need to prove that: $$54(2u+v)c^3-9v^2c^2+12(4u^3+6u^2v+4uv^2+v^3)c+2v^2(6u^2+6uv+v^2)\geq0,$$ which is true because $$54(2u+v)c^3-9v^2c^2+12(4u^3+6u^2v+4uv^2+v^3)c+2v^2(6u^2+6uv+v^2)\geq$$ $$\geq54vc^3-9v^2c^2+12v^3c+2v^4\geq0.$$

Answer 3

Hint: Let $x=a-c>0$ and $y=b-c>0$ and $z=3c\geq 0$, then we have to prove $${1\over 4x^2+z^2}+{1\over 4y^2+z^2}\geq {2\over (x+y+ z)^2},$$ which is obvious after full expanding.

Answer 4

A proof of the nonuser's inequality:

Let $x$, $y$ and $z$ be non-negative numbers such that $xy+xz+yz\neq0.$ Prove that: $${1\over4x^2+z^2}+{1\over4y^2+z^2}\geq{2\over(x+y+z)^2}.$$

Indeed, it's equivalent to $$2(x-y)^2(x^2+4x+y^2)+4xy(x+y)z-(x-y)^2z^2+2(x+y)z^3\geq0,$$ which is true by AM-GM $$2(x-y)^2(x^2+4x+y^2)+4xy(x+y)z-(x-y)^2z^2+2(x+y)z^3\geq$$ $$\geq2(x+y)z^3+(x-y)^2(x^2-2xy+y^2)-(x-y)^2z^2\geq$$ $$\geq3\sqrt[3]{(x+y)^2z^6\cdot(x-y)^4}-(x-y)^2z^2\geq$$ $$\geq\sqrt[3]{(x-y)^2z^6(x-y)^4}-(x-y)^2z^2=0.$$